Question
refer to the equation $A x^2+C y^2+D x+E y+F=0$, where $A$ and $C$ are not both 0 .Show that if $A=0, C \neq 0$, and $D \neq 0$, then the equation is equivalent to one of the form$$(y-k)^2=4 p(x-h)$$
Step 1
Since we are considering the case where \( A = 0 \), we can simplify the equation to: \[ C y^2 + D x + E y + F = 0 \] Show more…
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Suppose that $A$ and $C$ are not both $0 .$ Show that the set of all $(x, y)$ satisfying $$ A x^{2}+B x+C y^{2}+D y+E=0 $$ is either a parabola, an ellipse, or an hyperbola (or possibly $\emptyset$ ). Hint: The case $C=0$ is essentially Problem $15,$ and the case $A=0$ is just a minor variant. Now consider separately the cases where $A$ and $B$ are both positive or negative, and where one is positive while the other is negative.
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